1) đặt đk rùi bình phương 2 vế là ok

2) \(pt\Leftrightarrow\frac{\sqrt{x}-\sqrt{x+2}}{x-x-2}+\frac{\sqrt{x+2}-\sqrt{x+4}}{x+2-x-4}+\frac{\sqrt{x+4}-\sqrt{x+6}}{x+4-x-6}=\frac{\sqrt{10}}{2}-1\)(ĐKXĐ : \(x\ge0\))

<=> \(\frac{\sqrt{x}-\sqrt{x+6}}{-2}=\frac{\sqrt{10}}{2}-1\)

<=> \(\frac{\sqrt{x+6}-\sqrt{x}}{2}=\frac{\sqrt{10}-2}{2}\)

<=> \(\sqrt{x+6}-\sqrt{x}=\sqrt{10}-2\)

<=> \(\sqrt{x+6}+2=\sqrt{10}+\sqrt{x}\)

đến đây bình phương 2 vế rùi giải bình thường nhé 

\(\dfrac{-17}{15}\)

\(2\left(\sqrt{\frac{x^2+x+1}{x+4}}-1\right)+x^2-3=\frac{2}{\sqrt{x^2+1}}-1\)

\(\Leftrightarrow2\frac{\frac{x^2+x+1}{x+4}-1}{\sqrt{\frac{x^2+x+1}{x+4}}+1}+x^2-3=\frac{4-\left(x^2+1\right)}{\left(2+\sqrt{x^2+1}\right)\sqrt{x^2+1}}\)

\(\Leftrightarrow\frac{2\left(x^2-3\right)}{\sqrt{\left(x+4\right)\left(x^2+x+1\right)}+x+4}+x^2-3=\frac{3-x^2}{\left(2\sqrt{x^2+1}\right)\sqrt{x^2+1}}\)

\(\Leftrightarrow\left(x^2-3\right)\left(\frac{2}{\sqrt{\left(x+4\right)\left(x^2+x+1\right)}+x+4}+1+\frac{1}{\left(2+\sqrt{x^2+1}\right)\sqrt{x^2+1}}\right)=0\)

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(Cũng không chắc _-_ )

bạn làm đúng rồi đấy, mình đăng cho vuii thôi :)))

hello bạn

\(\Rightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{2x^3}{2}+\frac{x^2}{2}+\frac{2x}{2}+\frac{1}{2}\)

\(\Rightarrow\sqrt{x^2+x+\frac{1}{2}-\frac{1}{4}}=\sqrt{x^2+x+\frac{1}{4}}=x^3+\frac{x^2}{2}+x+\frac{1}{2}\)

\(\Rightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=x+\frac{1}{2}=x^3+\frac{x^2}{2}+x+\frac{1}{2}\)

\(\Rightarrow x^3+\frac{x^2}{2}+x+\frac{1}{2}-x-\frac{1}{2}=x^3+\frac{x^2}{2}=0\Rightarrow\frac{2x^3+x^2}{2}=0\)

\(\Rightarrow2x^3+x^2=0\Rightarrow x^2\left(2x+1\right)=0\Rightarrow\hept{\begin{cases}x^2=0\Rightarrow x=0\\2x+1=0\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\end{cases}}\)

vậy x=0 và x=-1/2

\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\) (*) (ĐKXĐ: \(\forall x\in R\))

\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[x^2\left(2x+1\right)+\left(2x+1\right)\right]\)

\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\left|x+\frac{1}{2}\right|}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)

+) Xét \(x+\frac{1}{2}\ge0\Leftrightarrow x\ge-\frac{1}{2}\). Khi đó pt (*) trở thành:

\(\sqrt{x^2-\frac{1}{4}+x+\frac{1}{2}}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)

\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)

\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\) (Do \(x\ge\frac{1}{2}\))

\(\Leftrightarrow\frac{\left(2x+1\right)\left(x^2+1\right)-\left(2x+1\right)}{2}=0\)

\(\Leftrightarrow x^2\left(2x+1\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\) (t/m ĐKXĐ)

+) Xét \(x+\frac{1}{2}< 0\Leftrightarrow x< -\frac{1}{2}\). Khi đó: \(2x+1< 0\)

Ta thấy: \(2x+1< 0;x^2+1>0;\frac{1}{2}>0\Rightarrow\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)< 0\)

Mà \(\sqrt{x^2-\frac{1}{4}+\left|x+\frac{1}{2}\right|}\ge0\) nên Vô lí ---> Loại TH này.

Vậy tập nghiệm của pt (*) là \(S=\left\{0;-\frac{1}{2}\right\}.\)

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